Evaluate the following: \( \begin{array}{llll}\text { (a) } i^{13} & \text { (b) } 2 i^{3} & \text { (c) } i^{7} & \text { (d) } i^{14}\end{array} \)

To evaluate the powers of \( i \), we first recall that \( i \) is the imaginary unit defined by the property \( i^2 = -1 \). The powers of \( i \) cycle every four terms: \[ \begin{align*} i^1 & = i, \\ i^2 & = -1, \\ i^3 & = -i, \\ i^4 & = 1, \end{align*} \] and then \( i^5 = i \), \( i^6 = -1 \), and so forth. Now, let's evaluate each term: (a) \( i^{13} \) To find \( i^{13} \), we can reduce the exponent modulo 4: \[ 13 \mod 4 = 1 \quad (\text{since } 13 = 3 \cdot 4 + 1) \] So, \( i^{13} = i^1 = i \). (b) \( 2i^{3} \) First, we evaluate \( i^3 \): \[ i^{3} = -i \] Thus, we have: \[ 2i^{3} = 2(-i) = -2i. \] (c) \( i^{7} \) To find \( i^{7} \), we reduce the exponent: \[ 7 \mod 4 = 3 \quad (\text{since } 7 = 1 \cdot 4 + 3) \] So, \( i^{7} = i^3 = -i \). (d) \( i^{14} \) For \( i^{14} \): \[ 14 \mod 4 = 2 \quad (\text{since } 14 = 3 \cdot 4 + 2) \] Thus, \( i^{14} = i^2 = -1 \). In summary, we have: - \( i^{13} = i \) - \( 2 i^{3} = -2i \) - \( i^{7} = -i \) - \( i^{14} = -1 \)